An inductor of 0 . 5 mH , a capacitor of 20 μF and resistance of 20 Ω are connected in series with a 220 V ac source. If the current is in phase with the emf, the amplitude of current of the circuit i

Question

An inductor of 0 . 5 mH , a capacitor of 20 μF and resistance of 20 Ω are connected in series with a 220 V ac source. If the current is in phase with the emf, the amplitude of current of the circuit is x A . The value of x is-

TestHub · Accepted Answer

Here, emf and current are in phase, so the circuit is either pure resistive circuit or case of resonance,

$X_{L} = X_{C}$

Thus, the impedance of circuit is $Z = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}} = R = 20 Ω$

Now, rms value of current is $i_{rms} = \frac{220}{20} = 11 A$

Using the relation between peak and rms value of current, we have peak value $i_{0 =} i_{rms} \sqrt{2} = 11 \sqrt{2} = \sqrt{242} A$ .

Hence, the value of $x = 242$ .

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